# Given scalene triangle ABC, what is 2bccosA+2accosB+2abcosC?

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should come up with the following notations for the sides of triangle such that: `BC=a, AC=b, AB=c` .

You need to use the law of cosines such that:

`BC^2 = AB^2 + AC^2 - 2AB*AC*cos hatA `

`BC^2- AB^2- AC^2 = - 2AB*AC*cos hatA`

`AB^2 + AC^2 - BC^2 = 2AB*AC*cos hatA `

`cos hatA = (AB^2 + AC^2 - BC^2)/(2AB*AC)`

You should substitute a for BC, b for AC, c for AB such that:

`cos hatA = (c^2 + b^2 - a^2)/(2c*b)`

`AB^2 = BC^2 + AC^2 - 2BC*AC*cos hatC`

`cos hatC = (BC^2 + AC^2 - AB^2)/(2BC*AC)`

`cos hatC = (a^2 + b^2 - c^2)/(2a*b)`

`AC^2 = AB^2 + BC^2 - 2BC*AB*cos hatB`

`cos hatB = (AB^2 + BC^2 - AC^2)/(2BC*AB)`

`cos hatB = (c^2 + a^2 - b^2)/(2a*c)`

You need to substitute  `(c^2 + b^2 - a^2)/(2c*b)`  for `cos hatA` , `(c^2 + a^2 - b^2)/(2a*c)`  for `cos hatB`  and `(a^2 + b^2 - c^2)/(2a*b)`  for `cos hatC`  such that:

`2bc*cos hat A + 2ac*cos hat B + 2ab*cos hat C = 2bc*(c^2 + b^2 - a^2)/(2c*b) + 2ac*(c^2 + a^2 - b^2)/(2a*c) + 2ab*(a^2 + b^2 - c^2)/(2a*b)`

`2bc*cos hat A + 2ac*cos hat B + 2ab*cos hat C = c^2 + b^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2`

Reducing like terms yields:

`2bc*cos hat A + 2ac*cos hat B + 2ab*cos hat C = a^2 + b^2 + c^2 Hence,`  evaluating the given expression yields `2bc*cos hat A + 2ac*cos hat B + 2ab*cos hat C = a^2 + b^2 + c^2` .

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